Alicense plate has six characters. three characters are letters, two characters are numbers (0-9), and one character is a letter or a number. the letters and numbers can repeat. note: there are 26 letters in the alphabet. how many different license plates can be made? answer plates what is the probability of getting a plate with abc123 or 123abc? p(abc123 or 123abc) = answer (enter as a reduced fraction using / for the fraction bar.)

1)Number of different license plates can be made =17576×100×260=456976000

2)Probability of getting a plate with abc123 or 123 abc=0.0000004

Given that :-A license plate has six characters.

As three characters are letters ∴ ways of these 3 letters into the plate with repetition = 26×26×26=17576 ways

and  two characters are numbers (0-9 - total 10 characters)∴ways of these 2 numbers into the plate with repetition=10×10 =100 ways

and one character is a letter or a number=26×10=260

So number of different license plates can be made =17576×100×260=456976000

Now Probability of getting a plate with abc123 or 123 abc

=P(abc123 or 123abc)=P(abc123)+P(123abc)-P(abc123)×P(123abc)=1/456976000+1/456976000-1/456976000×1/456976000

=1/456976000(1+1-1/456976000)

=1/456976000(2-0.0000002)=2/456976000=0.0000004

Probability of getting a plate with abc123 or 123 abc=0.0000004

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation:

what is the original number? you forgot to add it.